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The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle time

Puzzle time

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  • KlausK Offline
    KlausK Offline
    Klaus
    wrote on last edited by Klaus
    #6

    So you start with t = 0 (i.e. turn bulb 1 off), then t=1 (do nothing with bulb 2), then t=2 (turn bulb 3 of) and so forth. After the first round, all odd numbered bulbs are off and all even numbered ones are on.

    Correct so far?

    I haven't thought a lot about it yet, but the only fixed point seems to be that all bulbs are off. Hmm... My first guess is that it will return to that same state around time t=2^100.

    1 Reply Last reply
    • jon-nycJ Offline
      jon-nycJ Offline
      jon-nyc
      wrote on last edited by
      #7

      Your first paragraph is correct. You’re also correct that straight 0s is stable.

      Only non-witches get due process.

      • Cotton Mather, Salem Massachusetts, 1692
      1 Reply Last reply
      • jon-nycJ Offline
        jon-nycJ Offline
        jon-nyc
        wrote on last edited by jon-nyc
        #8

        My first guess was that it would return because lower numbers do. But you need to do better than guess.

        Like Ax I wrote code to play a bit but after 36 bulbs ran for a full day I gave up on the brute force method.

        Only non-witches get due process.

        • Cotton Mather, Salem Massachusetts, 1692
        AxtremusA 1 Reply Last reply
        • HoraceH Offline
          HoraceH Offline
          Horace
          wrote on last edited by
          #9

          Is it not inevitable that it returns to all off? I don’t think there can exist a loop which does not eventually return to every antecedent state. And if there is no loop then you have to assume that the infinity of future states must contain all off. Which makes a loop. Maybe the insight is that all such state machines are necessarily looping.

          Education is extremely important.

          1 Reply Last reply
          • HoraceH Offline
            HoraceH Offline
            Horace
            wrote on last edited by
            #10

            Pinhole loops are not possible because they would require two different antecedent states leading to the same next state. And it seems a pinhole loop is the only way the state would never revert to the initial conditions.

            Education is extremely important.

            1 Reply Last reply
            • jon-nycJ Offline
              jon-nycJ Offline
              jon-nyc
              wrote on last edited by
              #11

              Yes that is the answer. From reversibility it follows.

              (assuming you meant 'all on' not 'all off')

              Only non-witches get due process.

              • Cotton Mather, Salem Massachusetts, 1692
              1 Reply Last reply
              • HoraceH Offline
                HoraceH Offline
                Horace
                wrote on last edited by
                #12

                Whichever one is less racist, that’s the one I meant.

                Education is extremely important.

                1 Reply Last reply
                • jon-nycJ jon-nyc

                  My first guess was that it would return because lower numbers do. But you need to do better than guess.

                  Like Ax I wrote code to play a bit but after 36 bulbs ran for a full day I gave up on the brute force method.

                  AxtremusA Offline
                  AxtremusA Offline
                  Axtremus
                  wrote on last edited by
                  #13

                  @jon-nyc said in Puzzle time:

                  Like Ax I wrote code to play a bit but after 36 bulbs ran for a full day I gave up on the brute force method.

                  Just out of curiosity, I ran a simulation for 36 bulbs on a MacBook Air (1.6 GHz Intel Core i5). It took 10 minutes and 36 seconds to reach the "all on" state at the 22,839,252,821st iteration. I implemented my simulator using Go.

                  What did you use to implement and run your simulation?

                  1 Reply Last reply
                  • HoraceH Offline
                    HoraceH Offline
                    Horace
                    wrote on last edited by
                    #14

                    I would add the reversibility isn’t a sufficient condition to answer the problem, you also need a finite number of possible states.

                    Education is extremely important.

                    1 Reply Last reply
                    • jon-nycJ Offline
                      jon-nycJ Offline
                      jon-nyc
                      wrote on last edited by jon-nyc
                      #15

                      yes yes. there are fewer than 100*2^100 states.

                      Only non-witches get due process.

                      • Cotton Mather, Salem Massachusetts, 1692
                      1 Reply Last reply
                      • jon-nycJ Offline
                        jon-nycJ Offline
                        jon-nyc
                        wrote on last edited by
                        #16

                        Ax I was exaggerating but I used python on the Mac

                        Only non-witches get due process.

                        • Cotton Mather, Salem Massachusetts, 1692
                        1 Reply Last reply
                        • KlausK Offline
                          KlausK Offline
                          Klaus
                          wrote on last edited by Klaus
                          #17

                          Nice puzzle!

                          Let's analyze this some more.

                          Every state is in a cycle that has a length n. We just established that the set of states is the disjoint union of all cycles.

                          Let's consider a histogram of cycle lengths. For instance, "000....000" is in a cycle of length one, hence we add one to the bin for cycles of length 1.

                          What kind of shape will the histogram have? And the extra Nassim Taleb Black Swan question is: does it have a "fat tail"?

                          1 Reply Last reply
                          • jon-nycJ Offline
                            jon-nycJ Offline
                            jon-nyc
                            wrote on last edited by
                            #18

                            I graphed out some early ones, they do not. THey're in the general shape of a normal curve (not saying they're normal).

                            Only non-witches get due process.

                            • Cotton Mather, Salem Massachusetts, 1692
                            KlausK 1 Reply Last reply
                            • brendaB Offline
                              brendaB Offline
                              brenda
                              wrote on last edited by
                              #19

                              Are these Christmas tree bulbs? If so, the answer is that they will never all be on again until you buy new ones.

                              1 Reply Last reply
                              • jon-nycJ jon-nyc

                                I graphed out some early ones, they do not. THey're in the general shape of a normal curve (not saying they're normal).

                                KlausK Offline
                                KlausK Offline
                                Klaus
                                wrote on last edited by
                                #20

                                @jon-nyc what do you mean by "early ones"? . I was talking about cycle length distributions. I'm not sure we talk about the same thing.

                                1 Reply Last reply
                                • jon-nycJ Offline
                                  jon-nycJ Offline
                                  jon-nyc
                                  wrote on last edited by
                                  #21

                                  No we’re not. I graphed out the sum of lightbulbs that were on in each state change (every toggle changes the sum by +-1) so it moves around in a jerky continuous fashion). That formed a narrow bell curve.

                                  When I say early ones, I did it to the completion of the cycle, or even multiple cycles for smaller n. I did also graph them for n=100 but just for the first few million ticks.

                                  Only non-witches get due process.

                                  • Cotton Mather, Salem Massachusetts, 1692
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